Why Is Radioactive Decay Always Exponential?

[hunt_mat]

Why is radioactive decay always expoenetial?

 

[tom.stoer]

Let's assume we have N particles, where N will become a function of time N(t); let's assume that all these particles are independent from each other and that each particle does not know about its own history (there is no definition of "age"). In a time intervall dt we will find dN particles decaying via one specific decay channel (other decay channels not taken into account) where dN = const * N * dt. This can be integrated resulting in an exponential decay law.

I think that the above mentioned assumptions are sufficient to derive the exponential decay law.

 

[hunt_mat]

It is but you have (to my mind) still not explained why $$dN=\textrm{const}Ndt$$, I would say that:

$$dN=\frac{dN}{dt}dt$$

And I would expect tht either some other theory or experiment would tell me what dN/dt is.

 

[tom.stoer]

I think my assumptions (independence, no age) are sufficient to explain dN = const * N * dt. Isn't this simply Stochastic? What is missing?

 

[hunt_mat]

Forgive me, I know very little nuclear physics, I work in electrified fluids as a day job. Nothing may be missing but it's more of that I'm just not getting it perhaps, could you add in a little more explanation.

 

[tom.stoer]

Sorry, it's not that I don't want to ...

OK, at time t we have N(t) particles. Within time interval dt the number of particles changes (due to the decay) by dN. The probability that one specific particle decays is simply p = dN/N. This probability is constant, it does neither depend on N(t) nor on t.

It does not depend on t b/c an individal particle does not get older. If it would, p(t) would increase with time (the probability of dying for human beings does increase with age), but for particles there's no good reason for that.

And it does not depend on N b/c the particles are independent, they do not interact (there are situations where this indeed changes, e.g. for the neutron inside a nucleus; a free neutron deacays, whereas a neutron inside a nucleus is stable for many nuclides; this is not what we are talking about; we do not compare different conditions for our particles). The probability does not change if I add another particle to my ensemble of particles.

So p = dN/N, dN = p * N = (const * dt) * N. That means that if I have more particles, I will see more decays per time dt. It's the simplest equation I can write down, there is no good reason why it should not work, ... so let's see where does this leave us ...

... hope this helps

 

[hunt_mat]

Ah okay, this makes much more sense now, I see why p=dN/N is constant and that obviously in this model $$dN\propto dt$$, as in a longer time period you would expect more decays. I see how the model works now. Are there any corrections to this model?

 

[tom.stoer]

No corrections.

Ok, if the "environmental conditions" change, p changes as well. E.g. for a free neutron there's a specific p for beta decay. The constant p is different once you do not look at free neutrons but at neutrons which are bound inside a nucleus. Then for each nuclide you get a different p (it can even be zero), but again that does not change with time or with the number of nuclei.

 

[hunt_mat]

So this really is a very general theoretical law? Wow, I don't think that I have come across anything like this in the things I have worked in.

 

[petergreat]

So this really is a very general theoretical law? Wow, I don't think that I have come across anything like this in the things I have worked in.

It's very general because it arises from simple considerations. Consider a large number of bombs. Each bomb has a random number generator which, at every second, produces a new random integer in the range [1,10^4]. The bomb detonates itself once the number 1234 is produced. If we assume that the detonation of one bomb doesn't trigger other bombs, then it's easy to convince yourself that the number of remaining bombs decay exponentially, as Exp[-t/10^4], where t is the number of seconds elapsed. Of course, statistical variation will be present, which is also the case for real radioactivity.

In radioactive materials, you can think of every radioactive nucleus as a bomb described above. The assumption that the explosion of a bomb doesn't trigger others is a valid one concerning radioactivity. In other words, the decay of a nucleus doesn't trigger other nuclei to decay. (Otherwise it's called a "chain reaction" rather than radioactivity. Chain reactions are important for nuclear weapons.) The "random number generator" associated with each nucleus is from the "tunneling probability" in quantum mechanics that governs the behavior of subatomic particles.

 

[hunt_mat]

Thanks thisd has been very very helpful.

 

[sophiecentaur]

If the radioactive source is not just one radioactive substance then the decay will not be exponential. The recorded count will be the sum of counts due to two or more sets of decay statistics. The departure from exponential would depend on the relative levels of radiation from each element. If they are near-equal then that would be the worst case.

 

[hunt_mat]

So what would be the equation in that case?

 

[sophiecentaur]

The sum of two or more exponentials with different starting values and different half lives. Each decay, just as each nucleus, is independent of all else. Easy enough to write it out that way but a bit harder to determine the four coefficients if you're given the curve of the measured count values with time..

 

[hunt_mat]

Okay.

 

[Vanadium 50]

I'm a little surprised that you haven't seen the problem of A decaying to B, which decays into C etc. in your differential equations class. It's a classic problem - right up there with "tanks of brine". I bet if you look at your old textbook, you will find a set of problems of this sort there.

 

[hunt_mat]

I did very little applied problems like that, it was all of the case of, "This is how you solve an ODE of this form.." There was very little examples from these sciences and my first degree was in pure maths